Question

A gene can be either type A or type B, and it can be either dominant or recessive. If the gene is type B, then there is a probability of 0.31 that it is dominant. There is also a probability of 0.22 that a gene is type B and it is dominant.
What is the probability that a gene is of type A?

Answer 1

Answer: The probability that a gene is of type A is 0.29.

Step-by-step explanation:

Let A be the event of type A.

Let B be the event of type B.

Let D be the event of dominant.

So, P(B and D) = 0.22

P(D|B)=0.31

So, Using the conditional probability, we get that

$P(B|D)=\dfrac{P(B\cap D)}{P(B)}\\\\0.31=\dfrac{0.22}{P(B)}\\\\P(B)=\dfrac{0.22}{0.31}=0.71$

So, the probability that a gene is of type A is given by

$P(A)=1-P(B)=1-0.71=0.29$

Hence, the probability that a gene is of type A is 0.29.

Answer 2

Answer:

The probability that a gene is of type A is 0.29 .

Step-by-step explanation:

We are given that a gene can be either type A or type B, and it can be either dominant or recessive.

Let event A = gene having type A

     event B = gene having type B

     event D = gene is dominant

     event R = gene is recessive

We are given that if the gene is type B, then there is a probability of 0.31 that it is dominant i.e. P(D/B) = 0.31

Also, there is a probability of 0.22 that a gene is type B and it is dominant i.e. P(B $\bigcap$ D) = 0.22

Now, the conditional probability of P(D/B) is given by;

                  ⇒   P(D/B) = $\frac{P(B \bigcap D) }{P(B)}$  

                  ⇒  0.31 = $\frac{0.22 }{P(B)}$   ⇒ P(B) = 0.22 ÷ 0.31 = 0.71

So, now Probability that a gene is of type A, P(A) = 1 - P(B) {because only gene of type A or type B is possible}

            = 1 - 0.71 = 0.29 .

Hence, the probability that a gene is of type A is 0.29 .